# Prove that a group of order 3 must be cyclic.

Is my argument correct, if not what is wrong with it. If the group $G$ is not cyclic, this means that there is not element of order 3. So if $G=\{e, a, b\}$ then $|a|=2$ and $|b|=2$, a contradiction because by Lagrange's Theorem element order divide group order but 2 does not divide 3. Hence $G$ is cyclic.

I just want to show if this is correct or not and why, I do not need a proof.

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Yep, correct as written. – DonAntonio Nov 4 '12 at 13:36
@DonAntonio Thanks. – Reader Nov 4 '12 at 13:41
@Reader, have you had Cauchy's Theorem?: the order of an element must divide the order of the group. So if the group has order 3, then there must be an element of order 3 and hence powers of this element fill the whole group. – Nicky Hekster Nov 4 '12 at 15:15
@ Nicky Hekster Thanks. – Reader Nov 4 '12 at 15:30
The most elementary method is to use Sudoku! The products $ab$ and $ba$ must be equal to $e$ (using cancellation law). By default $a^2=b$. Therefore $a^3=a^2 a =ba=e$. – PAD Nov 4 '12 at 19:10